First and second laws of thermodynamics: getting increments from inexact differentials

Question

First law of thermodynamics states that $$\tag{1} dU=\delta Q + \delta W$$

whereas second law of thermodynamics states that, for a reversible process, $$\tag{2} dS= \frac{\delta Q}{T} $$

I have also seen $(1)$ and $(2)$ expressed in terms of variations, as follows:

$$\tag{3} \Delta U=\Delta Q + \Delta W$$

$$\tag{4} \Delta S= \frac{\Delta Q}{T} $$

In bothy equations $(1)$ and $2$ appears the inexact differential symbol, $\delta$, so I assume that those variables labelled with the $\delta$ need an integrating factor to be solved. However, it seems that the new equations have been achieved simply by performing an integral.

Then,

1. How are these equations transformed from their differential forms to their integral ones?

2. Are $(3)$ and $(4)$ as general $(1)$ and $(2)$, or they can be used only under certain conditions?

Answer 1

First law of thermodynamics states that 𝑑𝑈=𝛿𝑄+𝛿𝑊

The symbol 𝛿 is used instead of $d$ to represent inexact differentials because heat $Q$ and work $W$ are path dependent, whereas $d$ is the exact differential used for $U$ (internal energy) is not path dependent, i.e., it is a system property. According to Wikipedia the convention to use 𝛿 originated in the 19th century work of German mathematician Carl Gottried.

Although work and heat are path dependent, the combination of the two is not as they determine the change in internal energy, which is path independent. This is the reason the following form can be used.

$$\Delta U=Q+W$$

The more precise definition of entropy is

$$dS=\frac{𝛿Q_{rev}}{T}$$

Where the suffix $rev$ means heat transfer for a reversible path.

The form

$$\Delta S=\frac{Q}{T_{reservoir}}$$

Can be used when the heat transfer occurs at constant temperature, such as transfer with a heat reservoir, so that temperature can come out of the integral when integrating the differential form.

Hope this helps.